Problem: Simplify and expand the following expression: $ \dfrac{k}{k - 9}-\dfrac{2k}{k - 10} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(k - 9)(k - 10)$ Multiply the first term by $\dfrac{k - 10}{k - 10}$ $ \begin{align*} \dfrac{k}{k - 9} \times \dfrac{k - 10}{k - 10} & = \dfrac{(k)(k - 10)}{(k - 9)(k - 10)} \\ & = \dfrac{k^2 - 10k}{(k - 9)(k - 10)}\end{align*} $ Multiply the second term by $\dfrac{k - 9}{k - 9}$ $ \begin{align*} \dfrac{2k}{k - 10} \times \dfrac{k - 9}{k - 9} & = \dfrac{(2k)(k - 9)}{(k - 10)(k - 9)} \\ & = \dfrac{2k^2 - 18k}{(k - 10)(k - 9)}\end{align*} $ Now we have: $ = \dfrac{k^2 - 10k}{(k - 9)(k - 10)} - \dfrac{2k^2 - 18k}{(k - 10)(k - 9)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{k^2 - 10k - (2k^2 - 18k)}{(k - 9)(k - 10)} $ $ = \dfrac{k^2 - 10k - 2k^2 + 18k}{(k - 9)(k - 10)} $ $ = \dfrac{-k^2 + 8k}{(k - 9)(k - 10)}$ Expand the denominator: $ = \dfrac{-k^2 + 8k}{k^2 - 19k + 90}$